I may be wrong, but Orient Line/Line does not seem to snap to an intersect with a line?

- Brian

Hi Brian,

re:
> Does History work with Orient Line/Line?

it does if you use it with "Make copies" enabled and then select the generated object and use Edit > History > "Enable update".

- Michael

Maybe I over simplified it???

For an irregular polygon in a circle, the "Sum of all central angles is 2π".
The sum of the outer internal angles is 180 * (number of sides - 2). 180*(5-2) = 540 degrees.
So with the unknown being the rotation angle of the right unit chord,
the angle (~114.82 degrees) between two unit chords can be calculated, with a little simple math.

So devise a method to draw the right chord leg, and the right chord arm, with the changing rotation, (~97.75degrees), and snap the endpoint to top center, as the pair of chords is rotated.

https://math.stackexchange.com/questions/3184676/finding-the-interior-angles-of-an-irregular-polygon-inscribed-on-a-circle

- Brian

(114.82 * 3) + (97.76 * 2) = 539.98 = 540 degrees.
https://math.stackexchange.com/questions/3047152/if-the-total-number-of-angles-in-a-polygon-is-180n-2-why-are-there-just-36

Use radians instead of degrees...

3*phi + 2*theta = 3*PI with angles in radians

Simplifying, phi = PI - (2/3) * theta. (In radians)

phi is the inside angle, unit chord to unit chord.
theta is the inside angle between sqrt2 chord and connected unit chord.

Does History work with Dimension Angle. So that when an angle is increased by rotation, the Dimension Angle auto changes?
Is Dimension Angle Value available to some script or node? Does it have a radian display mode?
Is here a History manipulation Node?

I could not understand Burr's video. Is it a solution to the PentagonChord puzzle?

Thinking of MoI doing the Math, with a Snap to endpoint vs intersection or closed polyline, or ???, as the right leg is rotated, with dynamic angle update.

Or rotate pair of unit chords?
Or mirror unit chord pair, and check for closure to left endpoint of sqrt2 chord?
Rotation or mirror handle?

- Brian

Hi Elang,

I think it's impossible to have all sides touching the circle if one side is X√2 and other sides are x.
x are too short.
The answer would be a bird's house.

I scaled up by 10, for better visibility.
Base is now 10*sqrt(2).
Unit chords are now 10 units in length.

Initial MoI construction:
A right angle triangle, with angles 90, 45, 45, with the two perpendicular sides of length 10 units, has a hypotenuse of 10*sqrt(2).
Place the 10*sqrt(2) in the horizontal base position.
Place a future chord line segment of 10 units from the right end point, to the +x direction. It can be rotated to yield the right leg chord.
The inner angle theta is between 97.784758 degrees, and 97.7891905 degrees.
To use MoI's Rotate command, the angle is the supplement, (180 - theta) degrees. This rotation angle is about 82.2108095 degrees (plus a tiny amount).
((The rotation circle was pointed out by Michael.)

The angle between two scaled unit chords is Phi = 180 - (2/3) * theta.
Phi is (a little less than), approximately 114.8072063333 degrees.
To Rotate the right hand leg chord, to the right hand arm chord, use a Rotation angle of (360 - Phi), which equals (180 + (2/3) * theta).
So the angle to type in the Rotate box is (a little more than) 245.1927936667 degrees. (which is 114.x07... degrees on the other side of the angle.)

The independent variable is the angle theta, when rotating the right leg chord.
The three point circle, and the other 3 unit chords, and angle Phi, need to be updated during the rotation, and have a Snap, when the size of the circle, and the end points of the four 10*unit chords match up.

I need to learn more about History, and how to use it to do rotations, and 3 point circle, so that they will auto update, with theta angle change.

Or try to do a Node.




Alternate link:
https://imgbox.com/mfMcu1JY

The above image is a close approximation, but does not quite solve the problem.

- Brian

Manually doing trial Theta and Phi cord placement, to narrow down the values, takes too much time.

Play with trying to do History with 3 point circle, and History with two rotations, did not seem to work.

Hello all... Thank you very much for thoroughly replies.

I'm very sorry that I'm out of town for about a week (it's been 2 days till now), and I don't have access to Moi for this time being. So I'll keep reading your replies and will be continue to solve the task based on all of your replies immediately.

Cheers!

Dear BurrMan...

How do you define the circle radius/diameter at the very first place so it fits the categories?

Dear All...

Isn't x√2 is the diagonal of a square with x for its lengths? So... would it helps if the square added to the original diagram? Just a silly thinking...

Hi, Mip...

I believe the 'outer circle' is added at last time using 3 points circle.. And in your example, if you make the circle radius bigger, then the bird house with those lengths will be possible!

I believe here's the key:
1. Create a circle with B as center, and the radius should get through A2 and C... let's call this circle HELPER
2. Rotate this circle with A2 as pivot point, in particular ways, so that IF we draw 3 point circle from A1, A2, and B, then the circle WILL intersect HELPER along with the V Axis and C.

BUT, this will take several steps before we get the wanted result after zooming execisevely to the intersection point mentioned in number 2.

Just a crazey idea.

Hi,
Attached a nod file that allows manual adjustment to get what is needed.
Maybe possible to automate rather than eyeball ?
Cheers
Barry

Hi Elang,
I may have misunderstood the problem with an assumption error on my part

I thought the question was about orientation and how to rotate the items like magenta lines. so i made the video (nephew in school meant very young to me).

I didnt see massive math.

I have assumed that x√2 is a result of A2 to C?

The angle of x defined by a circle of radius x. Thats how i start with circle and orient.

If you have random angle of x, i don't think the problem solves.

If i have made an error in reading the initial post, i must drop off as it is much deeper than i am capable

“””””” Isn't x√2 is the diagonal of a square with x for its lengths? ””””

I used 2 triangles. C and B to Center of Circle and B and A2 to center of Circle. The diagonal of A2 to C

I dont think you are looking for 45-45-90

I think you want 60-60-60

That nephew must have access to a CAD program with constraint feature :)

x is equivalent to a scale factor. So x = 1 or 10, or even 100 is good.

The base chord, (x*sqrt(2)), can be generated as the hypotenuse of a 45 45 90 triangle with two sides of x*1 unit(s), by pythagoran.

The fact that 4 chords are of equal length x*1 units, implies that their central angles are equal, so the rotation angle from one to the other is equal,
with a value of approximately 114.81 degrees +/-. There are 3 of these equal angles, I am calling phi.

The angle at A2, of approximately 97.78, (I'm calling it theta), must be adjusted so that all vertices fit on the 3point circle.
Every time angle theta is adjusted, the angle phi also changes, by the simple formula previously given. So all of the x*unit chords keep changing as the candidate theta value changes. So they need to be redrawn over and over.
Constraining the 3 angles phi to a function of theta, would solve the puzzle, if the Cad program had constraints, and correct snap(s).

I guess a MoI script or a node could be written with an epsilon of the error of intersecting the centerline at C with chordBC?
Keep running the adjustment of theta, until epsilon reaches zero?

- Brian

My 2018 Alibre program is in a zip file.

I was able to run Barry's node program. (After replacing Display node. Either my copy or Barry's must be old?)
It draw's close approximations to the solution, subject to visual adjustment to approach a solution. (As far as I can tell.)

I do not have enough knowledge of MoI to go much further.

Trigonometry can replace phi with a function of theta, but the resulting equation would still likely require numerical methods to solve. (?)

It seems like people are solving the geometry kernels job.

But integrityware already does all that.

If you specify X or x√2 or the circle radius, then the rest is a few seconds in MoI.

Same if you want to calc from any given value of those 3 objects.

No need for constraints of any kind.

Unless you are trying to program automation of the process. Then i’ll step out and watch.

Hi, Barry.

This could be exactly what we need!... Thank you very much.
Now I just need to browse the forum to study how NOD works in Moi.

Cheers!

PS: This could be a prove that constraints is possible within Moi! How about integrates constrains to Moi, Michael? :)

Hi BurrMan,

- The horizontal line of A1-A2, with x√2 length comes first in the quiz.
- Followed by the 2 magenta lines with x length connected each other
- Arranged (move and rotate) the magenta lines in particular way
- ... so that if we draw 3 points circle (RED colored) through point A1,A2, and B, then it would also 'touch' C
- C should be on both Red Circle and V axis.

Cheers.